Rotational Kinematics
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AP Physics C: Mechanics › Rotational Kinematics
A wheel with moment of inertia $I=1.2\ \text{kg·m}^2$ must speed up from $\omega_0=5.0\ \text{rad/s}$ to $\omega_f=17\ \text{rad/s}$ in $t=4.0\ \text{s}$ under a constant net torque (friction included). Key equations: $\alpha=(\omega_f-\omega_0)/t$ and $\tau=I\alpha$. Using the given conditions, what is the torque required to achieve this change?
$\tau=1.8\ \text{N·m}$
$\tau=0.30\ \text{N·m}$
$\tau=14\ \text{N·m}$
$\tau=3.6\ \text{N·m}$
Explanation
This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating required torque for a desired angular acceleration. The problem requires working backwards from kinematic information to find the necessary torque. For a wheel with I = 1.2 kg·m² accelerating from ω0 = 5.0 rad/s to ωf = 17 rad/s in t = 4.0 s, we first find the angular acceleration. Choice B is correct because α = (ωf - ω0)/t = (17 - 5.0)/4.0 = 12/4.0 = 3.0 rad/s², and then τ = Iα = 1.2 × 3.0 = 3.6 N·m. Choice C at 14 N·m might result from calculation errors or using wrong values. To help students: emphasize the two-step process of finding acceleration first, then torque; practice problems that work backwards from desired motion to required forces/torques; and always verify that calculated values make physical sense.
A disk with moment of inertia $I=0.30\ \text{kg·m}^2$ is acted on by a constant torque $\tau=1.5\ \text{N·m}$ for $t=2.0\ \text{s}$, starting from rest; friction is negligible. Key equations: $\tau=I\alpha$ and $\theta=\tfrac12\alpha t^2$. Using the given conditions, calculate the total angular displacement during the 2.0 s interval.
$\theta=20\ \text{rad}$
$\theta=10\ \text{rad}$
$\theta=2.5\ \text{rad}$
$\theta=5.0\ \text{rad}$
Explanation
This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating angular displacement under constant angular acceleration. This parallels the translational kinematic equation for displacement under constant acceleration. For a disk with I = 0.30 kg·m² experiencing torque τ = 1.5 N·m for t = 2.0 s starting from rest, we need to find total angular displacement. Choice A is correct because first we find α = τ/I = 1.5/0.30 = 5.0 rad/s², then use θ = ½αt² = ½(5.0)(2.0)² = ½(5.0)(4.0) = 10 rad. Choice C at 20 rad might result from forgetting the factor of ½ in the kinematic equation. To help students: draw parallels between rotational and translational kinematic equations; emphasize that the ½ factor comes from integration just as in linear motion; and practice problems starting from rest to simplify calculations.
A gyroscope has spin angular momentum magnitude $L=0.40\ \text{kg·m}^2/\text{s}$. Gravity exerts a torque of magnitude $\tau=0.080\ \text{N·m}$ about the pivot, producing steady precession; assume the spin magnitude stays constant. Key equation: $\Omega=\tau/L$. Using the given conditions, determine the precession rate $\Omega$.
$\Omega=2.0\ \text{rad/s}$
$\Omega=5.0\ \text{rad/s}$
$\Omega=0.032\ \text{rad/s}$
$\Omega=0.20\ \text{rad/s}$
Explanation
This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically gyroscopic precession. Precession occurs when a torque acts perpendicular to a spinning object's angular momentum, causing the spin axis to rotate. For a gyroscope with angular momentum L = 0.40 kg·m²/s experiencing torque τ = 0.080 N·m, the precession rate is found using the gyroscopic equation. Choice A is correct because Ω = τ/L = 0.080/0.40 = 0.20 rad/s. Choice B at 5.0 rad/s would result from inverting the fraction or other calculation errors. To help students: explain that precession is different from ordinary rotation; emphasize that the precession equation Ω = τ/L applies when spin angular momentum is much larger than precession angular momentum; and use demonstrations or videos to visualize gyroscopic motion.